Question

An experiment is performed on a sample of atoms known to have a ground state of −5.0 eV. The gas is illuminated with “white light” (400 – 700 nm). A spectrometer capable of analyzing radiation in this range is used to measure the radiation. The sample is observed to absorb light at only 400 nm. After the “white light” is turned off, the sample is observed to emit visible radiation of 400 nm and 600 nm.

(a) In the space below, determine the values of the energy levels and on the following scale sketch an energy level diagram showing the energy values in eV’s and the relative positions of:

i. the ground state

ii. the energy level to which the system was first excited

iii. one other energy level that the experiment suggests may exist

(b) What is the wavelength of any other radiation, if any, that might have been emitted in the experiment? Why was it not observed?

Answer 1

(a) i. The ground state was given in the problem. 

ii. The first excited state corresponds to the absorption of the 400 nm photon. This energy is given by E = hc / λ … 1240/400 = 3.1 eV. Moving from a state of –5 eV and absorbing 3.1 eV moves you up to the higher excited state of –1.9 eV. 

iii. After reaching the excited state, the electron makes two visible drops. It drops back down the ground state emitting the 400 nm photon, or it emits the 600 nm photon. This photon corresponds to an energy of 2.1 eV (found with E = hc / λ again). Starting at –1.9eV and emitting 2.1 eV would put you in the –4eV level.

(b) From the diagram above, the emission not yet analyzed is from the ‘other’ state to the ground state. That emission corresponds to an energy release of 1 eV. The λ of this emission can be found with E = hc / λ. Giving λ = 1240 nm. Since this is outside of the visible spectrum, it is not ‘seen’, but it does occur.